If $f(x)=x^3+e^\frac{x}{2}$, compute $\frac{d}{dx} \left(f^{-1}(x)\right) \text{at$~~x=1$}$

Question

If $f(x)=x^3+e^\frac{x}{2}$ We need to compute $$\frac{d}{dx} \left(f^{-1}(x)\right) \text{at$~~x=1$}$$ I tried to find $f^{-1}(x)$, but i couldn't

A hint that i got said If $f(x)$ and $g(x)$ are inverse of each other then $$f^{'}(\alpha).g^{'}(\beta)=1$$

Where $(\alpha,\beta)$ is any point on $f(x)$

Now this did actually help me get the answer but how can we prove the above condition given as a Hint

All suggestions to deduce the condition are welcome

Answer 1

I very well know that you can find the answer to your question (a derivative of the inverse function) But I guess you are looking for the explanation for why $f'(\alpha)\times g'(\beta) = 1$ where $(\alpha, \beta) $ is point on $y = f(x)$

At first, let me tell you I'm bad at maths what I'll be explaining is just what I could imagine

• What is an inverse function:

If $y = f(x)$ is some function then the function obtained by changing the Y-axis with X-axis is inverse.

• Why $f'\times g' = 1$ The tangent to $y = f(x) $ must be perpendicular to tangent of $f^{-1}(x)$ Now, If the slope of the tangent to function$y = f(x)$ is $f'(x)_{\alpha}$ then the slope of the tangent to its inverse function must be equal to reciprocal ($-\frac 1{f'(x)_{x = g(x)}}$)However, your view has been changed(rotated by 90-degrees) negative sign will not be there;

Thus,

$$g'(x) = \frac d{dx}f^{-1}(x) = \frac 1{f'(g(x))} = \frac 1{f'(f^{-1}(x))}$$

Answer 2

Note that $f(0)=1$. (Also, $f$ is strictly increasing on $(0,\infty)$ so it is one-to-one there).

Hence, $f^{-1}(1)=0$ and $(\frac d {dx} f^{-1}(x))_{x=1}=\frac 1 {f'(f^{-1}(1))}=\frac 1 {f'(0)}=2$.